Question

How much heat is needed to melt 16.6 kg of lead that is initially at 19.5°C?

Answer 1

To melt this amount of lead, you need to heat it to its melting point and then supply enough heat to liquify the lead

the heat needed to warm lead to its melting point is:

Q=m*c*ΔT where Q is the heat needed, m is the mass, c the specific heat and ΔT is the temperature change

here: m=16.6kg, c=127.6J/kg/K, and ΔT=327.5 - 19.5 = 308 degree celsius
so that ΔT=308+273= 581 K (a change of one degree C is the same as a change of one degree K, so it is OK to say this ΔT is 581 K)
Q=m*c*ΔT = 16.6*127.6*581= 1.230x10^6J
so for this part of the process, Q=1.230x10^6J

the second part requires the heat to melt 16.6 kg of lead, and this is just the mass of the sample x its heat of fusion, so the amount of heat needed here is 16.6kgx24.5x10^3J/kg or
4.06x10^5J

the total heat needed is 1.230x10^6J +4.06x10^5J  

=1.636x10^6J